A 16Ω wire is bend to form a square loop. A 9V battery with internal resistance 1Ω is connected across one of its sides.

Question

A \(16\Omega\) wire is bend to form a square loop. A 9V battery with internal resistance \(1\Omega\) is connected across one of its sides. If a \(4\mu F\) capacitor is connected across one of its diagonals, the energy stored by the capacitor will be \(\frac{x}{2} \mu J\). where x = _________.

Answer 1

x = 81

\(I = \cfrac{V}{R_{eq}} = \cfrac{9}{1+\cfrac{12\times4}{12+4}} = \cfrac{9}{4}\)

\(I_1 = \frac{9}{4} \times \frac{4}{16} = \frac{9}{16}\)

\(V_A - V_B = I_1 \times 8 = \frac{9}{16} \times 8 = \frac{9}{2} V\)

\(\therefore U = \frac{1}{2} \times 4 \times \frac{81}{4} \mu J\)

\(\therefore U = \frac{81}{2} \mu J\)

\(\therefore x = 81\)